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**Unformatted text preview: **cos2 (t) + 41
= 1. Solving, we
x
get cos(t) = ± 16x2 −1
16x2 √ =± 16x2 −1
. Since t belongs to − π , 0
2
4√ |
|x
16−x2
= 4|x| . (The absolute values here ∪ 0, π , we know
2 cos(t) ≥ 0, so we choose cos(t)
are necessary, since
x could be negative.) Since the domain of arccsc(x) requires |x| ≥ 1, the domain of
arccsc(4x) requires |4x| ≥ 1. Using Theorem 2.3, we can rewrite this as the compound
1
inequality 4x ≤ −1 or 4x ≥ 1. Solving, we get x ≤ − 1 or x ≥ 4 . Since we had no
4
√ additional restrictions on t, the equivalence cos(arccsc(4x)) =
1
1
−∞, − 4 ∪ 4 , ∞ . 16x2 −1
4 |x | holds for all x in 10.6 The Inverse Trigonometric Functions 10.6.2 711 Inverses of Secant and Cosecant: Calculus Friendly Approach π
In this subsection, we restrict f (x) = sec(x) to 0, π ∪ π , 32
2 y y 3π
2 1
π
π
2 π 3π
2 x −1 π
2 reﬂect across y = x f (x) = sec(x) on 0, π
2 ∪ π, −− − − − −→
−−−−−− 3π
2 switch x and y...

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